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1 year ago

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A boy drags a suitcase along the ground with a force of 100 N. If the frictional force opposing the motion of the suitcase is 50

N, what is the resultant forward force on the suitcase?

1 answer:

stira [4]1 year ago

4 0

Fortunately, 'force' is a vector. So if you know the strength and direction
of each force, you can easily addum up and find the 'resultant' (net) force.

When we talk in vectors, one newton forward is the negative of
one newton backward. Hold that thought, while I slog through
the complete solution of the problem.

(100 N forward) plus (50 N backward)

= (100 N forward) minus (50 N forward)

= 50 N forward .

That's it.
Is there any part of the solution that's not clear ?

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Two parallel plates are a distance apart with a potential difference between them. a point charge moves from the negatively char

Ghella [55]

Answer:

We can relate the kinetic energy of the particle to the potential difference between the plates by following equations:

Work energy theorem:

$W_{total} = \Delta K = K_2 - K_1 = w - 0$

$W = Fx\\F = W/d = w/d$

$F = Eq \\E = F/q = w/(dq)$

$V = Ed = \frac{w}{dq} d = w/q$

So,

$w = Vq$

If the distance is doubled and the potential difference is halved, then

$w = qV/2$

Explanation:

As can be seen from the relationship between kinetic energy and the potential difference, the distance between the plates has no effect on the relation between kinetic energy and the potential difference. Since the charge of the second particle is equal to that of the first one, the new kinetic energy would be half of the first kinetic energy.

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Sandy is on a road trip. She leaves at 8:00 AM. It takes her 2 hours to drive 200 kilometers. She stops at a rest stop for half

lutik1710 [3]

The average velocity of Sandy is given by the total distance covered S divided by the total time taken t:
$v= \frac{S}{t}$

The total distance covered is
$S=200 km+0+100 km=300 km$
while the total time taken is 2 hours + half an hour (for the rest) + 1 hour and half, so
$t=2h+0.5h+1.5 h=4 h$
Therefore, the average velocity is
$v= \frac{S}{t}= \frac{300 km}{4 h}=75 km/h$

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1 year ago

A metal sphere of radius 2.0 cm carries an excess charge of 3.0 μC. What is the electric field 6.0 cm from the center of the sph

Nonamiya [84]

Answer:

The electric field is $E = 7.5 *10^{6} \ N/C$

Explanation:

From the question we are told that

The radius of the metal sphere is $R = 2.0 \ cm = 0.02 \ m$

The excess charge which the metal sphere carries is $q = 3.0 \mu C = 3.0*10^{-6} \ C$

The distance of the position being to the center is $D = 6.0 \ cm = 0.06 \ m$

The coulomb constant is $k =9*10^{9} \ N \cdot m^2 /C^2$

Generally the electric field is mathematically represented as

$E = \frac{k * q}{D^2}$

substituting values

$E = \frac{9*10^{9} * 30.*10^{-6}}{(0.06)^2}$

$E = 7.5 *10^{6} \ N/C$

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568 muons were counted by a detector on the top of Mount Washington in a one hour period of time. Assuming moving muons keep tim

AlladinOne [14]

The answer to this question is:

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1 year ago

Read 2 more answers

A thin insulating rod is bent into a semicircular arc of radius a, and a total electric charge Q is distributed uniformly along

storchak [24]

Answer:

$v = \frac{kQ}{a}$

Explanation:

We define the linear density of charge as:

$\lambda = \frac{Q}{L}$

Where L is the rod's length, in this case the semicircle's length L = πr

The potential created at the center by an differential element of charge is:

$dv = \frac{kdq}{r}$

where k is the coulomb's constant

r is the distance from dq to center of the circle

Thus.

$v = \int_{}^{}\frac{kdq}{a}$

$v = \frac{k}{a}\int_{}^{}dq$

$v = \frac{kQ}{a}$ Potential at the center of the semicircle

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1 year ago